Is there a list of common identities for principal branches of complex logarithms and roots?

Question

Often I have a necessity to simplify some expressions involving roots or logarithms, where arguments appear to be imaginary or even more generally complex. Due to $n$th root and logarithm being generally multivalued, many simple identities, which hold for positive arguments, no longer hold for negative or complex ones.

Thus, for example, when I have an expression like

$$\frac{\sqrt{i\pi+x}}{\sqrt{-i\pi-x}},$$

I have to stop and analyze, whether I really can "extract $-1$" from a radical, and which sign the resulting $i$ will have — taking into account how the principal branch of the multivalued function (square root here) is defined.

So, is there any list of exactly working (i.e. not "up to branch") identities for principal branches of roots and logarithms, which would help simplification of such expressions?

Answer 1

There are no useful identities in the strictest sense of the word "identity". But if you know some a priori information on the range of operands, you may be able to use it to simplify your expressions with logarithms and roots. Below are listed some relations for this. They are also useful for having a feel when one may expect that an expression can be simplified, and when it's not really easy.

1. Notation

Whenever we use relations of $<$, $>$, $\le$, $\ge$, this implies that both sides of such relations are real-valued. Variables will be denoted as follows:

• $i=\sqrt{-1}$
• $r\in\mathbb R$,
• $p>0$,
• $z=x+iy\in\mathbb C$,
• $\alpha\in(-1,1]$.

A variable may have an index to make different variables of the same type distinct, e.g. both $p_1$ and $p_2$ are positive.

2. Logarithm

We denote multivalued logarithm and complex argument functions by $\ln$ and $\arg$ respectively. Corresponding principal branches of these functions are denoted by $\newcommand{\Ln}{\operatorname{Ln}}\newcommand{\Arg}{\operatorname{Arg}} \Ln$ and $\Arg$. The principal branch of $\Arg$ is defined by $$-\pi<\Arg(z)\le\pi,$$ and $\Ln$ is by definition $$\Ln(z)=\ln|z|+i\Arg(z).$$

2.1. Logarithm of scaled by a positive constant complex number

This is just an extension of the commonly known identity for all-positive arguments: $$\Ln(pz)=\Ln(p)+\Ln(z)$$

2.2. Logarithm of phase-shifted complex argument

$$\Ln((-1)^\alpha z)=\begin{cases} \Ln(z)+\alpha\pi i, &\text{if }-(\alpha+1)\pi<\Arg(z)\le(1-\alpha)\pi,\\ \Ln(z)+(\alpha-2)\pi i, &\text{if }\Arg(z)>(1-\alpha)\pi,\\ \Ln(z)+(\alpha+2)\pi i, &\text{if }\Arg(z)\le-(\alpha+1)\pi. \end{cases}$$ Particular cases of this are $\Ln(-z)$ and $\Ln(\pm iz)$, corresponding to $\alpha=0,\pm\frac12$: $$ \Ln(-z)=\begin{cases} \Ln(z)-\pi i, &\text{if }\Arg(z)>0,\\ \Ln(z)+\pi i, &\text{if }\Arg(z)\le0. \end{cases} $$ $$ \Ln(iz)=\begin{cases} \Ln(z)-\frac{3\pi}2, &\text{if }\Arg(z)>\frac\pi2,\\ \Ln(z)+\frac\pi2, &\text{if }\Arg(z)\le\frac\pi2. \end{cases} $$ $$ \Ln(-iz)=\begin{cases} \Ln(z)+\frac{3\pi}2, &\text{if }\Arg(z)\le-\frac\pi2,\\ \Ln(z)-\frac\pi2, &\text{if }\Arg(z)>-\frac\pi2. \end{cases} $$

2.3. Logarithm of product of a complex and a real

$$ \Ln(rz)=\begin{cases} \Ln(r)+\Ln(z), &\text{if }(\Arg(z)\le0\text{ or }r>0),\\ \Ln(r)+\Ln(z)-2\pi i, &\text{if }(\Arg(z)>0\text{ and }r<0). \end{cases} $$

2.4. Sum of logarithms of complex numbers

$$ \Ln(z_1)+\Ln(z_2)=\begin{cases} \Ln(z_1z_2), &\text{if }|\Arg(z_1)+\Arg(z_2)|\le\pi,\\ \Ln(z_1z_2)+\pi i, &\text{if }\pi<\Arg(z_1)+\Arg(z_2)<2\pi,\\ \ln(z_1z_2)+2\pi i, &\text{if }(z_1<0\text{ and }z_2<0),\\ \Ln(z_1z_2)-\pi i, &\text{if }\Arg(z_1)+\Arg(z_2)<-\pi. \end{cases} $$

2.5. Logarithm scaled by a real number

$$ -\pi<r\Arg(z)\le\pi\implies r\Ln(z)=\Ln(z^r). $$ In particular, for the range-restricted version of $r$, which we denote by $\alpha$, $$ \alpha\Ln(z)=\Ln(z^\alpha). $$

2.6. Negated logarithm of a complex number

$$ -\Ln(z)=\begin{cases} \Ln\left(\frac1z\right), &\text{if }|\Arg(z)|<\pi,\\ \Ln\left(\frac1z\right)-2\pi i &\text{if }z<0. \end{cases} $$

3. Square root

Principal branch of square root is defined as $$ \sqrt{z}=\exp\left(\frac i2\Arg(z)\right)\sqrt{|z|} $$

3.1. Square root of scaled by a positive constant complex number

Similarly to logarithm, square root has a nice property for product of a complex number and a positive number: $$ \sqrt{pz}=\sqrt p\sqrt z. $$

3.2. Square root of phase-shifted complex number

$$ \sqrt{(-1)^\alpha z}=\begin{cases} \sqrt z \exp\left(\frac{i\pi}2(\alpha-2)\right), &\text{if }\Arg(z)>(1-\alpha)\pi,\\ \sqrt z \exp\left(\frac{i\pi\alpha}2\right), &\text{if }-(\alpha+1)\pi<\Arg(z)\le(1-\alpha)\pi,\\ \sqrt z \exp\left(\frac{i\pi}2(\alpha+2)\right), &\text{if }\Arg(z)\le-(1+\alpha)\pi. \end{cases} $$

Particular cases of this are $\sqrt{-z}$ and $\sqrt{\pm iz}$, corresponding to $\alpha=0,\pm\frac12$:

$$ \sqrt{-z}=\begin{cases} -i\sqrt z, &\text{if }\Arg(z)>0,\\ i\sqrt z, &\text{if }\Arg(z)\le0. \end{cases} $$ $$ \sqrt{iz}=\begin{cases} \frac{-1-i}{\sqrt2}\sqrt z, &\text{if }\Arg(z)>\frac\pi2,\\ \frac{1+i}{\sqrt2}\sqrt z, &\text{if }\Arg(z)\le\frac\pi2, \end{cases} $$ $$ \sqrt{-iz}=\begin{cases} \frac{1-i}{\sqrt2}\sqrt z, &\text{if }\Arg(z)>-\frac\pi2,\\ \frac{-1+i}{\sqrt2}\sqrt z, &\text{if }\Arg(z)\le-\frac\pi2. \end{cases} $$

3.3. Square root of product of a complex and a real

$$ \sqrt{rz}=\begin{cases} \sqrt r\sqrt z, &\text{if }(\Arg(z)>0\text{ or }r\ge0),\\ -\sqrt r\sqrt z, &\text{if }(\Arg(z)\le0\text{ and }r<0). \end{cases} $$

3.4. Product of square roots of complex numbers

$$ \sqrt{z_1}\sqrt{z_2}=\begin{cases} \sqrt{z_1z_2}, &\text{if }-\pi<\Arg(z_1)+\Arg(z_2)\le\pi,\\ -\sqrt{z_1z_2}, &\text{if }\Arg(z_1)+\Arg(z_2)\in(-\infty,-\pi]\cup(\pi,\infty). \end{cases} $$

Answer 2

There can't really be any such identities written down in a useful way.

The common ways of writing down things results in expressions that define analytic functions, so if we have an identity $f(z)=g(z)$, the expression $f(z)-g(z)$ will define an analytic function that is identically zero in the domain where the identity is valid. But by the identity theorem this means that our identity ought to keep holding as we move $z$ smoothly around the branch points. Which means that the identity doesn't really take principal branches into account after all.

We can cheat this argument by picking uncommon ways to write down $f$ and $g$ (or by choosing a sufficiently uninteresting identity such as $\operatorname{Log}z = \operatorname{Log}z$) -- but generally that would mean that using the identity will itself require the same kind of painstaking analysis you're hoping to get out of.